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pychu [463]
3 years ago
5

A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of

her shadow changing when she is 15 ft from the base of the light?
Physics
2 answers:
Soloha48 [4]3 years ago
7 0

Answer:

The length of her shadow is changing at the rate  -2 m/s

Explanation:

Let the height oh the street light, h = 22 ft

Let the height of the woman, w = 5.5 ft

Horizontal distance to the street light = l

length of shadow = x

h/w = (l + x)/x

22/5.5 =  (l + x)/x

4x = l + x

3x = l

x = 1/3 l

taking the derivative with respect to t of both sides

dx/dt = 1/3 dl/dt

dl/dt = -6 ft/sec ( since the woman is walking towards the street light, the value of l is decreasing with time)

dx/dt = 1/3 * (-6)

dx/dt = -2 m/s

sveticcg [70]3 years ago
5 0

Answer:

Explanation:

By applying similar triangles rule,

22/5.5 = (x + s)/s

s = 5.5/22 × (x + s)

ds/dt = 5.5/22 × (dx/dt + ds/dt)

16.5/22 × ds/dt = 5.5/22 × dx/dt

16.5/22 × ds/dt = 5.5/22 × -6

ds/st = -2 ft/s.

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grin007 [14]

Answer:

A positive velocity

Explanation:

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3 years ago
How can you use distance and displacement to describe an object's motion​
andrew11 [14]

Answer:

distance can describe the total distance moved and displacement shows how far something has moved from its starting position (in a straight line from point a to point b) the object doesn't have to move in a straight line, but that is how displacement is measured

8 0
3 years ago
An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V  .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0
3 years ago
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
3 years ago
A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the
ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

8 0
3 years ago
Read 2 more answers
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